How do we show that the Gaussian integral$\displaystyle{\large{{\int_{{-\infty}}^{{\infty}}}{{e}}^{{-{{x}}^{{2}}}}{d}{x}}}$ converges?
I can figure out the convergence for $\displaystyle{\large{{\int_{{0}}^{{\infty}}}{{e}}^{{-{{x}}^{{2}}}}{d}{x}}}$ as follows:
We know that $e^{-x^2} \leq e^{-x}$ when $x \geq 1$ and then following this we get
$\displaystyle{\large{{\int_{{0}}^{{\infty}}}{{e}}^{{-{{x}}^{{2}}}}{d}{x}={\int_{{0}}^{{1}}}{{e}}^{{-{{x}}^{{2}}}}{d}{x}+{\int_{{1}}^{{\infty}}}{{e}}^{{-{{x}}^{{2}}}}{d}{x}}}$
and for $\displaystyle{\large{{\int_{{1}}^{{\infty}}}{{e}}^{{-{x}}}{d}{x}=\lim_{{{s}\to\infty}}{\int_{{1}}^{{s}}}{{e}}^{{-{x}}}{d}{x}=\lim_{{{s}\to\infty}}{\left(-{{e}}^{{-{x}}}{{\mid}_{{1}}^{{s}}}\right)}=\lim_{{{s}\to\infty}}{\left(-{{e}}^{{-{s}}}+{{e}}^{{-{1}}}\right)}=\frac{{1}}{{e}}}}$
so from the Comparison test, this shows that
$\displaystyle{\large{{\int_{{0}}^{{\infty}}}{{e}}^{{-{{x}}^{{2}}}}{d}{x}}}$ converges also, but Im confused with the one where the lower limit is $-\infty$.
Substituting $x \to -x$ yields: $$ \int_{-\infty}^0 e^{-x^2} \; \mathrm{d}x = \int_{\infty}^0 e^{-(-x)^2}(-1) \; \mathrm{d} x= \int_0^{\infty} e^{-x^2} \; \mathrm{d}x $$ Therefore: \begin{align*} \int_{-\infty}^{\infty} e^{-x^2} \; \mathrm{d}x &= \int_{-\infty}^0 e^{-x^2} \; \mathrm{d}x + \int_0^{\infty} e^{-x^2} \; \mathrm{d}x \\ &= \int_0^{\infty} e^{-x^2} \; \mathrm{d}x + \int_0^{\infty} e^{-x^2} \; \mathrm{d}x \\ &= 2\int_0^{\infty} e^{-x^2} \; \mathrm{d}x \\ &< \infty \end{align*}